What is the tangent of (4x+5)^5,\at x=-1 ?

The tangent of (4x+5)^5,\at x=-1 is y=20x+21

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tangent of (4x+5)^5,\at x=-1

tangent of

(4 x + 5)^{5}, at x = - 1

y = 20 x + 21

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

y = (4 x + 5)^{5} : \frac{d y}{d x} = 20 (4 x + 5)^{4}

Compute the slope of

y = (4 x + 5)^{5}

x = - 1 : m = 20

Find the line with slope m=

20

and passing through

(- 1, 1) : y = 20 x + 21

y = 20 x + 21

\frac{d}{d x} (\frac{e ^{x}}{8 x ^{2} + 1})

csc (y) d x + sec (x) d y = 0

f (x) = e^{2 x} arcsin (x)

\frac{d}{d x} (10 - x^{2})

y = x^{3} - 9 x

What is the tangent of (4x+5)^5,\at x=-1 ?
The tangent of (4x+5)^5,\at x=-1 is y=20x+21